c - 访问冲突读取位置 0x00000000。与 argv[]

Question

我正在运行以下程序并遇到错误:

First-chance exception at 0x0f32d440 (msvcr100d.dll) in c.exe: 0xC0000005: Access violation reading location 0x00000000.
Unhandled exception at 0x772815de in c.exe: 0xC0000005: Access violation reading location 0x00000000.
The program '[9048] c.exe: Native' has exited with code -1073741510 (0xc000013a).

这是代码

#include <string.h>
#include <stdio.h> 
#include <stdlib.h> 
int main(int argc, char *argv[], char *env[]) //char *argv[] 
{ 
int i; 

printf("These are the %d command- line arguments passed to main:\n\n", argc); 
if(strcmp(argv[1],"123")==0)                
    {
        printf("success\n");
     }
else
for(i=0; i<=argc; i++) 
    //if(strcmp(argv[1],"abc")==0)

    printf("argv[%d]:%s\n", i, argv[i]);
/*printf("\nThe environment string(s)on this system are:\n\n"); 
for(i=0; env[i]!=NULL; i++) 
printf(" env[%d]:%s\n", i, env[i]);*/ 
system("pause");
} 

问题应该出在strcmp函数上，但我不知道如何解决。谁能帮忙？

Answer 1

你(至少)有两个问题。

第一个是这样做的:

if(strcmp(argv[1],"123")==0) 

没有首先检查 argc >= 2。

第二个是这样的:

for(i=0; i<=argc; i++) 

因为您应该处理参数 0 到 argc - 1(含)。该循环所做的是通过 argc 处理参数 0 并且 argv[argc] 始终为 NULL。

以下程序说明了解决此问题的一种方法:

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int main (int argc, char *argv[]) {
    int i;

    printf ("These are the %d command-line argument(s) passed to main:\n", argc);
    if ((argc >= 2) && (strcmp (argv[1], "123") == 0)) {
        printf ("   success\n");
    } else {
        for (i = 0; i < argc; i++)  {
            printf ("   argv[%d] = [%s]\n", i, argv[i]);
        }
    }
    return 0;
}

您可以看到，只有在确保正确填充了 argv[1] 之后，才会与 "123" 进行比较。此外，循环已更改为排除 argv[argc]，因为这不是参数之一。成绩单如下:

pax> testprog
These are the 1 command-line argument(s) passed to main:
   argv[0] = [testprog]

pax> testprog 123
These are the 2 command-line argument(s) passed to main:
   success

pax> testprog a b c
These are the 4 command-line argument(s) passed to main:
   argv[0] = [testprog]
   argv[1] = [a]
   argv[2] = [b]
   argv[3] = [c]